A gas of 0.20\,mol occupies 5.010^-3\,m^3 at 300\,K. Find its pressure. Take R = 8.31\,J mol^-1K^-1.

WJEC WJEC 2019-style practice: A sealed container of volume 0.025\,m^3 holds an ideal gas at a pressure of 1.2 × 10^5\,Pa and temperature 290\,K. Calculate the number of moles of gas, and determine the mean kinetic energy of a molecule. Take R = 8.31\,J mol^-1\,K^-1 and k = 1.38 × 10^-23\,J K^-1.

Number of moles from pV = nRT: n = pVRT = 1.2 × 10^5 × 0.0258.31 × 290 = 30002410 = 1.24\,mol. Mean kinetic energy of a molecule from 12mc^2 = 32kT: E_k = 32kT = 32 × 1.38 × 10^-23 × 290 = 6.0 × 10^-21\,J. Markers reward the moles from the ideal gas equation and the mean kinetic energy from 32kT.

WalesPhysicsSyllabus dot point

How do the gas laws and kinetic theory connect the pressure of a gas to the motion of its molecules?

The gas laws and the ideal gas equation, the kinetic theory derivation, and the link between molecular kinetic energy and temperature.

A focused answer to WJEC A-Level Physics Unit 3 ideal gases, covering the gas laws and the ideal gas equation, the kinetic theory derivation of pV, and the link between the mean kinetic energy of a molecule and absolute temperature.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

WJEC wants you to use the gas laws and the ideal gas equation, outline the kinetic theory derivation of pressure, and link the mean kinetic energy of a molecule to absolute temperature. This dot point is where the macroscopic gas laws meet the microscopic motion of molecules, giving temperature a clear physical meaning, and the examiners reward students who can move fluently between the two pictures.

The answer

The gas laws and ideal gas equation

The gas laws (Boyle's, Charles's and the pressure law) combine into the ideal gas equation. Use $n$ (moles) with the molar gas constant $R$ , or the number of molecules $N$ with the Boltzmann constant $k$ , where $R = N_A k$ .

Kinetic theory

The kinetic model assumes many identical molecules in random motion, of negligible volume, making perfectly elastic collisions with no intermolecular forces between collisions. The derivation gives:

Temperature and molecular energy

Comparing $pV = \tfrac{1}{3}Nm\overline{c^2}$ with $pV = NkT$ gives:

Root mean square speed of nitrogen

A nitrogen molecule has mass $4.7 \times 10^{-26}\,\text{kg}$ . Find its root mean square speed at $300\,\text{K}$ . Take $k = 1.38 \times 10^{-23}\,\text{J K}^{-1}$ .

step Equate molecular kinetic energy to temperature

$\tfrac{1}{2}m\overline{c^2} = \tfrac{3}{2}kT$ , so $\overline{c^2} = \dfrac{3kT}{m}$ .

step Substitute

$\overline{c^2} = \dfrac{3 \times 1.38 \times 10^{-23} \times 300}{4.7 \times 10^{-26}} = \dfrac{1.24 \times 10^{-20}}{4.7 \times 10^{-26}} = 2.64 \times 10^{5}\,\text{m}^2\,\text{s}^{-2}$ .

step Take the square root

$c_{\text{rms}} = \sqrt{2.64 \times 10^{5}} = 514\,\text{m s}^{-1}$ , faster than the speed of sound, which is consistent because sound travels through this fast random motion.

Examples in context

Example 1. A car tyre warming on a motorway: As a tyre heats up, the gas inside obeys $pV = nRT$ . With the volume nearly fixed, the rising temperature raises the pressure, which is why tyre pressures are quoted "cold" and rise after a long drive. Kinetic theory explains it as faster molecules hitting the walls harder and more often.
Example 2. A weather balloon expanding: As a balloon rises, the outside pressure falls. At roughly constant temperature, Boyle's law $pV = \text{constant}$ means the volume expands as the pressure drops, so the balloon swells until, high in the atmosphere, it may burst. The same ideal-gas equation predicts the final volume from the pressure ratio.
Example 3. Why helium balloons go cold when filled: When gas rushes out of a high-pressure cylinder to fill a balloon, it expands rapidly and does work pushing back the surrounding air, so its internal energy and temperature fall. Kinetic theory links this temperature drop to the molecules slowing down as they spread out. The ideal-gas picture, where temperature reflects the mean molecular kinetic energy, explains the chilled feel of a freshly filled balloon.

Try this

Q1. A gas of $0.20\,\text{mol}$ occupies $5.0\times10^{-3}\,\text{m}^3$ at $300\,\text{K}$ . Find its pressure. Take $R = 8.31\,\text{J mol}^{-1}\text{K}^{-1}$ . [3 marks]

Cue. $p = \dfrac{nRT}{V} = \dfrac{0.20\times8.31\times300}{5.0\times10^{-3}} \approx 1.0\times10^5\,\text{Pa}$ .

Q2. State how the mean kinetic energy of a gas molecule depends on temperature. [1 mark]

Cue. It is directly proportional to the absolute temperature.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20195 marksA sealed container of volume

0.025\,\text{m}^3

holds an ideal gas at a pressure of

1.2 \times 10^{5}\,\text{Pa}

and temperature

290\,\text{K}

. Calculate the number of moles of gas, and determine the mean kinetic energy of a molecule. Take

R = 8.31\,\text{J mol}^{-1}\,\text{K}^{-1}

and

k = 1.38 \times 10^{-23}\,\text{J K}^{-1}

Show worked answer →

Number of moles from $pV = nRT$ :

$n = \dfrac{pV}{RT} = \dfrac{1.2 \times 10^{5} \times 0.025}{8.31 \times 290} = \dfrac{3000}{2410} = 1.24\,\text{mol}$ .

Mean kinetic energy of a molecule from $\tfrac{1}{2}m\overline{c^2} = \tfrac{3}{2}kT$ :

$E_k = \tfrac{3}{2}kT = \tfrac{3}{2} \times 1.38 \times 10^{-23} \times 290 = 6.0 \times 10^{-21}\,\text{J}$ .

Markers reward the moles from the ideal gas equation and the mean kinetic energy from $\tfrac{3}{2}kT$ .

WJEC 20214 marksShow that, for a fixed mass of ideal gas at constant temperature, doubling the volume halves the pressure, and explain this using kinetic theory.

Show worked answer →

At constant temperature and fixed amount of gas, $pV = nRT$ has $nRT$ constant, so $pV = \text{constant}$ (Boyle's law). If the volume doubles to $2V$ , then to keep the product constant the pressure must become $p/2$ , so the pressure halves.

Kinetic theory explanation: pressure arises from molecules colliding with the container walls. Doubling the volume means each molecule travels further between wall collisions, so the rate of collisions per unit area falls. With the temperature unchanged the molecules' speeds are unchanged, so the reduced collision rate halves the pressure. Markers reward the Boyle's law argument from $pV = nRT$ and the collision-rate explanation.

Related dot points

Sources & how we know this

WJEC A-level Physics specification — WJEC (2015)